prove an inequality between the curvature of these curves The exercise is: 
Let $\alpha$ a plane curve such that $|\alpha'(s)|=1$ with curvature $k(s)$. Let $\beta(s)=\alpha(s) + k(s)N(s)$ such that $\beta'(s)\ne 0$ $\forall s$.
($N$ is the normal vector of $\alpha$).
Is is true that $|k_{\beta}(s)| \le |k(s)|$ ? 
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The problem is that I don't know how to start doing it. The first thing I did was to work out the equality $k_{\beta}(s) = \lt T'_{\beta}(s),N_{\beta}(s)\gt$ to finally compare with $k(s)$ and solve the problem, but this lead me to a like 30 addends formula which I can't work with, and this makes me think that this is not the way I have to deal with this exercise, and maybe I just need to understand geometrically or intuitivelly the solution but I don't know how.
 A: It is not true. I present a counter-example.
Let $\alpha (s) = 2(\cos(s/2), \sin (s/2))$ be an arc-lenght parameterisation of the circle of radius $2$ centered at the origin ($s$ ranges from $0$ to $4\pi$). Then, $\alpha'(s) = (-\sin (s/2), \cos (s/2))$; since $s$ is an arc-length parameter for $\alpha$, then $T_\alpha(s) = \alpha'(s)$. The second derivative shall be $\alpha''(s) = \frac{1}{2}(-\cos(s/2), \sin (s/2))$. Note that $N_\alpha(s)$, which is the result of applying a $90$-degree rotation to $T_\alpha(s)$ is $N_\alpha(s)=(-\cos(s/2),\sin(s/2))$. Observe now that $$k(s) = <T'_\alpha, N_\alpha(s)> = <\alpha''(s),N_\alpha(s)> = <\frac{1}{2}N_\alpha(s),N_\alpha(s)> = 1/2.$$
So the scalar curvature of $\alpha$, $k(s)$, is constantly $1/2$, so we have that $|k(s)|= 1/2$.
Now let's go with $\beta(s) = \alpha(s) + k(s)N_\alpha(s)$. Since $\beta(s) = 1/2$ and $N_\alpha = (-\cos (s/2),-\sin(s/2))$, we have $$\beta(s) = \frac{3}{2}(\cos(s/2), \sin(s/2)),$$
which is the circle of radius $3/2$ ($s$ still ranges from $0$ to $4\pi$). Observe that this parameterisation is not arc-length, and we shall reparameterise it with
$$\beta(u) = \frac{3}{2}\left(\cos \frac{2u}{3}, \sin\frac{2u}{3}\right),
$$
with $u$ ranging from $0$ to $3\pi$. Going through the same calculations we made above for $\alpha(s)$ should yield that $k_\beta(u) = 2/3$.
Since $|k_\beta| = 2/3 > 1/2 = |k|$, the statement in your question is false.
